All categories

3 years ago

Solve for a.

Mathematics

2 answers:

Paraphin [41]3 years ago

3 0

C. All real numbers

cricket20 [7]3 years ago

3 0

Distribute it would look like this 4a plus 12 = 12 + 4a then you would subtract 4a from both sides and you would get 12 = 12

You might be interested in

Y is six more than x divided by three turn it into an expression (y=mx+b format)

Snowcat [4.5K]

We always start from right side in order of operation in any equation

we can set up equation as

step-1:

x divided by three

we can write as

$=\frac{x}{3}$

step-2:

six more than x divided by three

so, we get

$=\frac{x}{3}+6$

step-3:

y is six more than x divided by three

so, we get

$y=\frac{x}{3}+6$

and this is in y=mx+b form

so, we have

$y=\frac{x}{3}+6$ ...............Answer

4 0

3 years ago

If f(x)=-x+8 and g(x)=x^4,what is (g•f)(2)?

wariber [46]

Answer:

It can be written as f(−8) or f(3(−2)−2)

Explanation:

You would substitute −2 for the x in 3x−2 and then insert 3(−2)−2 for the g. You would end up with f(3(−2)−2), which can also be simplified to f(−8)

5 0

3 years ago

a jar contains 30 red, 40 blue, and 50 white buttons. you pick one button at random, find the probability that it is red or blue

abruzzese [7]

Answer:

0.58 rounded to the nearest hundredths

Step-by-step explanation:

30+40+50=120

now add red+blue

30+40=70

70/120

pls mark me brainliest

5 0

3 years ago

Read 2 more answers

Find two vectors in R2 with Euclidian Norm 1 whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .